Largest subarray with equal number of 0s and 1s

Given an array containing only 0s and 1s, find the largest subarray which contains equal no of 0s and 1s. Expected time complexity is O(n).

Examples:

Input: arr[] = {1, 0, 1, 1, 1, 0, 0}
Output: 1 to 6 
(Starting and Ending indexes of output subarray)

Input: arr[] = {1, 1, 1, 1}
Output: No such subarray

Input: arr[] = {0, 0, 1, 1, 0}
Output: 0 to 3 Or 1 to 4

Method 1: Brute Force.

Approach: The brute force approach in these type of questions is to generate all the possible sub-arrays. Then firstly check whether the sub-array has equal number of 0’s and 1’s or not. To make this process easy take cumulative sum of the sub-arrays taking 0’s as -1 and 1’s as it is. The point where cumulative sum = 0 will signify that the sub-array from starting till that point has equal number of 0’s and 1’s. Now as this is a valid sub-array, compare it’s size with the maximum size of such sub-array found till now.

Algorithm :



  1. Use a starting a pointer which signifies the starting point of the sub-array.
  2. Take a variable sum=0 which will take the cumulative sum of all the sub-array elements.
  3. Initialise it with value 1 if value at starting point=1 else initialise it with -1.
  4. Now start an inner loop and start taking cumulative sum of elements following same logic.
  5. If the cumulative sum (value of sum)=0 it signifies that sub-array has equal number of 0’s and 1’s.
  6. Now compare it’s size with size of the largest sub-array, if it is greater store the first index of such sub-array in a variable and update value of size.
  7. Print the sub-array with starting index and size returned by the above algorithm.

Pseudo Code:

Run a loop from i=0 to n-1
  if(arr[i]==1)
  sum=1
  else
  sum=-1
  Run inner loop from j=i+1 to n-1
      sum+=arr[j]
      if(sum==0)
        if(j-i+1>max_size)
           start_index=i
           max_size=j-i+1
Run a loop from i=start_index till max_size-1
print(arr[i])    

C++

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// A simple C++ program to find the largest
// subarray with equal number of 0s and 1s
#include <bits/stdc++.h>
  
using namespace std;
  
// This function Prints the starting and ending
// indexes of the largest subarray with equal
// number of 0s and 1s. Also returns the size
// of such subarray.
  
int findSubArray(int arr[], int n)
{
    int sum = 0;
    int maxsize = -1, startindex;
  
    // Pick a starting point as i
    for (int i = 0; i < n - 1; i++) {
        sum = (arr[i] == 0) ? -1 : 1;
  
        // Consider all subarrays starting from i
        for (int j = i + 1; j < n; j++) {
            (arr[j] == 0) ? (sum += -1) : (sum += 1);
  
            // If this is a 0 sum subarray, then
            // compare it with maximum size subarray
            // calculated so far
            if (sum == 0 && maxsize < j - i + 1) {
                maxsize = j - i + 1;
                startindex = i;
            }
        }
    }
    if (maxsize == -1)
        cout << "No such subarray";
    else
        cout << startindex << " to "
             << startindex + maxsize - 1;
  
    return maxsize;
}
  
/* Driver code*/
int main()
{
    int arr[] = { 1, 0, 0, 1, 0, 1, 1 };
    int size = sizeof(arr) / sizeof(arr[0]);
  
    findSubArray(arr, size);
    return 0;
}
  
// This code is contributed by rathbhupendra

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C

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// A simple program to find the largest subarray
// with equal number of 0s and 1s
  
#include <stdio.h>
  
// This function Prints the starting and ending
// indexes of the largest subarray with equal
// number of 0s and 1s. Also returns the size
// of such subarray.
  
int findSubArray(int arr[], int n)
{
    int sum = 0;
    int maxsize = -1, startindex;
  
    // Pick a starting point as i
  
    for (int i = 0; i < n - 1; i++) {
        sum = (arr[i] == 0) ? -1 : 1;
  
        // Consider all subarrays starting from i
  
        for (int j = i + 1; j < n; j++) {
            (arr[j] == 0) ? (sum += -1) : (sum += 1);
  
            // If this is a 0 sum subarray, then
            // compare it with maximum size subarray
            // calculated so far
  
            if (sum == 0 && maxsize < j - i + 1) {
                maxsize = j - i + 1;
                startindex = i;
            }
        }
    }
    if (maxsize == -1)
        printf("No such subarray");
    else
        printf("%d to %d", startindex, startindex + maxsize - 1);
  
    return maxsize;
}
  
/* Driver program to test above functions*/
  
int main()
{
    int arr[] = { 1, 0, 0, 1, 0, 1, 1 };
    int size = sizeof(arr) / sizeof(arr[0]);
  
    findSubArray(arr, size);
    return 0;
}

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Java

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class LargestSubArray {
  
    // This function Prints the starting and ending
    // indexes of the largest subarray with equal
    // number of 0s and 1s. Also returns the size
    // of such subarray.
  
    int findSubArray(int arr[], int n)
    {
        int sum = 0;
        int maxsize = -1, startindex = 0;
        int endindex = 0;
  
        // Pick a starting point as i
  
        for (int i = 0; i < n - 1; i++) {
            sum = (arr[i] == 0) ? -1 : 1;
  
            // Consider all subarrays starting from i
  
            for (int j = i + 1; j < n; j++) {
                if (arr[j] == 0)
                    sum += -1;
                else
                    sum += 1;
  
                // If this is a 0 sum subarray, then
                // compare it with maximum size subarray
                // calculated so far
  
                if (sum == 0 && maxsize < j - i + 1) {
                    maxsize = j - i + 1;
                    startindex = i;
                }
            }
        }
        endindex = startindex + maxsize - 1;
        if (maxsize == -1)
            System.out.println("No such subarray");
        else
            System.out.println(startindex + " to " + endindex);
  
        return maxsize;
    }
  
    /* Driver program to test the above functions */
  
    public static void main(String[] args)
    {
        LargestSubArray sub;
        sub = new LargestSubArray();
        int arr[] = { 1, 0, 0, 1, 0, 1, 1 };
        int size = arr.length;
  
        sub.findSubArray(arr, size);
    }
}

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Python3

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# A simple program to find the largest subarray
# with equal number of 0s and 1s
  
# This function Prints the starting and ending
# indexes of the largest subarray with equal 
# number of 0s and 1s. Also returns the size 
# of such subarray.
def findSubArray(arr, n):
  
    sum = 0
    maxsize = -1
  
    # Pick a starting point as i
  
    for i in range(0, n-1):
      
        sum = -1 if(arr[i] == 0) else 1
  
        # Consider all subarrays starting from i
  
        for j in range(i + 1, n):
          
            sum = sum + (-1) if (arr[j] == 0) else sum + 1
  
            # If this is a 0 sum subarray, then 
            # compare it with maximum size subarray
            # calculated so far
  
            if (sum == 0 and maxsize < j-i + 1):
                  
                maxsize = j - i + 1
                startindex = i
              
          
      
    if (maxsize == -1):
        print("No such subarray");
    else:
        print(startindex, "to", startindex + maxsize-1);
  
    return maxsize
  
# Driver program to test above functions
arr = [1, 0, 0, 1, 0, 1, 1]
size = len(arr)
findSubArray(arr, size)
  
# This code is contributed by Smitha Dinesh Semwal

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// A simple program to find the largest subarray
// with equal number of 0s and 1s
using System;
  
class GFG {
  
    // This function Prints the starting and ending
    // indexes of the largest subarray with equal
    // number of 0s and 1s. Also returns the size
    // of such subarray.
  
    static int findSubArray(int[] arr, int n)
    {
        int sum = 0;
        int maxsize = -1, startindex = 0;
        int endindex = 0;
  
        // Pick a starting point as i
        for (int i = 0; i < n - 1; i++) {
            sum = (arr[i] == 0) ? -1 : 1;
  
            // Consider all subarrays starting from i
  
            for (int j = i + 1; j < n; j++) {
                if (arr[j] == 0)
                    sum += -1;
                else
                    sum += 1;
  
                // If this is a 0 sum subarray, then
                // compare it with maximum size subarray
                // calculated so far
  
                if (sum == 0 && maxsize < j - i + 1) {
                    maxsize = j - i + 1;
                    startindex = i;
                }
            }
        }
        endindex = startindex + maxsize - 1;
        if (maxsize == -1)
            Console.WriteLine("No such subarray");
        else
            Console.WriteLine(startindex + " to " + endindex);
  
        return maxsize;
    }
  
    // Driver program
    public static void Main()
    {
  
        int[] arr = { 1, 0, 0, 1, 0, 1, 1 };
        int size = arr.Length;
        findSubArray(arr, size);
    }
}
  
// This code is contributed by Sam007

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<?php 
// A simple program to find the 
// largest subarray with equal 
// number of 0s and 1s
  
// This function Prints the starting 
// and ending indexes of the largest 
// subarray with equal number of 0s 
// and 1s. Also returns the size of 
// such subarray.
function findSubArray(&$arr, $n)
{
    $sum = 0;
    $maxsize = -1;
  
    // Pick a starting point as i
  
    for ($i = 0; $i < $n - 1; $i++)
    {
        $sum = ($arr[$i] == 0) ? -1 : 1;
  
        // Consider all subarrays
        // starting from i
        for ($j = $i + 1; $j < $n; $j++)
        {
            ($arr[$j] == 0) ? 
               ($sum += -1) : ($sum += 1);
  
            // If this is a 0 sum subarray, 
            // then compare it with maximum 
            // size subarray calculated so far
  
            if ($sum == 0 && $maxsize < $j - $i + 1)
            {
                $maxsize = $j - $i + 1;
                $startindex = $i;
            }
        }
    }
    if ($maxsize == -1)
        echo "No such subarray";
    else
        echo $startindex. " to " .
            ($startindex + $maxsize - 1);
  
    return $maxsize;
}
  
// Driver Code
$arr = array(1, 0, 0, 1, 0, 1, 1);
$size = sizeof($arr);
  
findSubArray($arr, $size);
  
// This coed is contributed 
// by ChitraNayal
?>

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Output:

 0 to 5

Complexity Analysis:

  • Time Complexity: O(n^2).
    As all the possible sub-arrays are generated using a pair of nested loops.
  • Auxiliary Space: O(1).
    As no extra data structure is used which takes auxiliary space.

Method 2: Hashmap.

Approach: The concept of taking cumulative sum, taking 0’s as -1 will help us in optimising the approach. While taking the cumulative sum, there are two cases when there can be a sub-array with equal number of 0’s and 1’s.

  1. When cumulative sum=0, which signifies that sub-array from index (0) till present index has equal number of 0’s and 1’s.
  2. When we encounter a cumulative sum value which we have already encountered before, which means that sub-array from the previous index+1 till the present index has equal number of 0’s and 1’s as they give a cumulative sum of 0 .

In a nutshell this problem is equivalent to finding two indexes i & j in array[] such that array[i] = array[j] and (j-i) is maximum. To store the first occurrence of each unique cumulative sum value we use a hash_map wherein if we get that value again we can find the sub-array size and compare it with maximum size found till now.

Algorithm :

  1. Let input array be arr[] of size n and max_size be the size of output sub-array.
  2. Create a temporary array sumleft[] of size n. Store the sum of all elements from arr[0] to arr[i] in sumleft[i].
  3. There are two cases, the output sub-array may start from 0th index or may start from some other index. We will return the max of the values obtained by two cases.
  4. To find the maximum length sub-array starting from 0th index, scan the sumleft[] and find the maximum i where sumleft[i] = 0.
  5. Now, we need to find the subarray where subarray sum is 0 and start index is not 0. This problem is equivalent to finding two indexes i & j in sumleft[] such that sumleft[i] = sumleft[j] and j-i is maximum. To solve this, we create a hash table with size = max-min+1 where min is the minimum value in the sumleft[] and max is the maximum value in the sumleft[]. Hash the leftmost occurrences of all different values in sumleft[]. The size of hash is chosen as max-min+1 because there can be these many different possible values in sumleft[]. Initialize all values in hash as -1.
  6. To fill and use hash[], traverse sumleft[] from 0 to n-1. If a value is not present in hash[], then store its index in hash. If the value is present, then calculate the difference of current index of sumleft[] and previously stored value in hash[]. If this difference is more than maxsize, then update the maxsize.
  7. To handle corner cases (all 1s and all 0s), we initialize maxsize as -1. If the maxsize remains -1, then print there is no such subarray.

Pseudo Code:

int sum_left[n]
Run a loop from i=0 to n-1
  if(arr[i]==0)
  sumleft[i] = sumleft[i-1]+-1
  else
  sumleft[i] = sumleft[i-1]+-1
        if (sumleft[i]  max)
            max = sumleft[i];


Run a loop from i=0 to n-1
 if (sumleft[i] == 0)
        {
           maxsize = i+1;
           startindex = 0;
        }
 
        // Case 2: fill hash table value. If already
        then use it

        if (hash[sumleft[i]-min] == -1)
            hash[sumleft[i]-min] = i;
        else
        {
            if ((i - hash[sumleft[i]-min]) > maxsize)
            {
                maxsize = i - hash[sumleft[i]-min];
                startindex = hash[sumleft[i]-min] + 1;
            }
        }

return maxsize

C

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// A O(n) program to find the largest subarray
// with equal number of 0s and 1s
  
#include <stdio.h>
#include <stdlib.h>
  
// A utility function to get maximum of two
// integers
  
int max(int a, int b) { return a > b ? a : b; }
  
// This function Prints the starting and ending
// indexes of the largest subarray with equal
// number of 0s and 1s. Also returns the size
// of such subarray.
  
int findSubArray(int arr[], int n)
{
    // variables to store result values
  
    int maxsize = -1, startindex;
  
    // Create an auxiliary array sunmleft[].
    // sumleft[i] will be sum of array
    // elements from arr[0] to arr[i]
  
    int sumleft[n];
  
    // For min and max values in sumleft[]
  
    int min, max;
    int i;
  
    // Fill sumleft array and get min and max
    // values in it.  Consider 0 values in arr[]
    // as -1
  
    sumleft[0] = ((arr[0] == 0) ? -1 : 1);
    min = arr[0];
    max = arr[0];
    for (i = 1; i < n; i++) {
        sumleft[i] = sumleft[i - 1]
                     + ((arr[i] == 0) ? -1 : 1);
        if (sumleft[i] < min)
            min = sumleft[i];
        if (sumleft[i] > max)
            max = sumleft[i];
    }
  
    // Now calculate the max value of j - i such
    // that sumleft[i] = sumleft[j]. The idea is
    // to create a hash table to store indexes of all
    // visited values.
    // If you see a value again, that it is a case of
    // sumleft[i] = sumleft[j]. Check if this j-i is
    // more than maxsize.
    // The optimum size of hash will be max-min+1 as
    // these many different values of sumleft[i] are
    // possible. Since we use optimum size, we need
    // to shift all values in sumleft[] by min before
    // using them as an index in hash[].
  
    int hash[max - min + 1];
  
    // Initialize hash table
  
    for (i = 0; i < max - min + 1; i++)
        hash[i] = -1;
  
    for (i = 0; i < n; i++) {
        // Case 1: when the subarray starts from
        // index 0
  
        if (sumleft[i] == 0) {
            maxsize = i + 1;
            startindex = 0;
        }
  
        // Case 2: fill hash table value. If already
        // filled, then use it
  
        if (hash[sumleft[i] - min] == -1)
            hash[sumleft[i] - min] = i;
        else {
            if ((i - hash[sumleft[i] - min]) > maxsize) {
                maxsize = i - hash[sumleft[i] - min];
                startindex = hash[sumleft[i] - min] + 1;
            }
        }
    }
    if (maxsize == -1)
        printf("No such subarray");
    else
        printf("%d to %d", startindex,
               startindex + maxsize - 1);
  
    return maxsize;
}
  
/* Driver program to test above functions */
int main()
{
    int arr[] = { 1, 0, 0, 1, 0, 1, 1 };
    int size = sizeof(arr) / sizeof(arr[0]);
  
    findSubArray(arr, size);
    return 0;
}

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C++/STL

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// C++ program to find largest subarray with equal number of
// 0's and 1's.
  
#include <bits/stdc++.h>
using namespace std;
  
// Returns largest subarray with equal number of 0s and 1s
  
int maxLen(int arr[], int n)
{
    // Creates an empty hashMap hM
  
    unordered_map<int, int> hM;
  
    int sum = 0; // Initialize sum of elements
    int max_len = 0; // Initialize result
    int ending_index = -1;
  
    for (int i = 0; i < n; i++)
        arr[i] = (arr[i] == 0) ? -1 : 1;
  
    // Traverse through the given array
  
    for (int i = 0; i < n; i++) {
        // Add current element to sum
  
        sum += arr[i];
  
        // To handle sum=0 at last index
  
        if (sum == 0) {
            max_len = i + 1;
            ending_index = i;
        }
  
        // If this sum is seen before, then update max_len
        // if required
  
        if (hM.find(sum + n) != hM.end()) {
            if (max_len < i - hM[sum + n]) {
                max_len = i - hM[sum + n];
                ending_index = i;
            }
        }
        else // Else put this sum in hash table
            hM[sum + n] = i;
    }
  
    for (int i = 0; i < n; i++)
        arr[i] = (arr[i] == -1) ? 0 : 1;
  
    printf("%d to %d\n",
           ending_index - max_len + 1, ending_index);
  
    return max_len;
}
  
// Driver method
  
int main()
{
    int arr[] = { 1, 0, 0, 1, 0, 1, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    maxLen(arr, n);
    return 0;
}
  
// This code is contributed by Aditya Goel

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Java

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import java.util.HashMap;
  
class LargestSubArray1 {
  
    // Returns largest subarray with
    // equal number of 0s and 1s
  
    int maxLen(int arr[], int n)
    {
        // Creates an empty hashMap hM
  
        HashMap<Integer, Integer> hM
            = new HashMap<Integer, Integer>();
  
        // Initialize sum of elements
        int sum = 0;
  
        // Initialize result
        int max_len = 0;
        int ending_index = -1;
        int start_index = 0;
  
        for (int i = 0; i < n; i++) {
            arr[i] = (arr[i] == 0) ? -1 : 1;
        }
  
        // Traverse through the given array
  
        for (int i = 0; i < n; i++) {
            // Add current element to sum
  
            sum += arr[i];
  
            // To handle sum=0 at last index
  
            if (sum == 0) {
                max_len = i + 1;
                ending_index = i;
            }
  
            // If this sum is seen before,
            // then update max_len if required
            if (hM.containsKey(sum + n)) {
                if (max_len < i - hM.get(sum + n)) {
                    max_len = i - hM.get(sum + n);
                    ending_index = i;
                }
            }
            else // Else put this sum in hash table
                hM.put(sum + n, i);
        }
  
        for (int i = 0; i < n; i++) {
            arr[i] = (arr[i] == -1) ? 0 : 1;
        }
  
        int end = ending_index - max_len + 1;
        System.out.println(end + " to " + ending_index);
  
        return max_len;
    }
  
    /* Driver program to test the above functions */
    public static void main(String[] args)
    {
        LargestSubArray1 sub = new LargestSubArray1();
        int arr[] = { 1, 0, 0, 1, 0, 1, 1 };
        int n = arr.length;
  
        sub.maxLen(arr, n);
    }
}
  
// This code has been by Mayank Jaiswal(mayank_24)

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# Python 3 program to find largest 
# subarray with equal number of 
# 0's and 1's. 
  
# Returns largest subarray with 
# equal number of 0s and 1s 
def maxLen(arr, n): 
  
    # NOTE: Dictonary in python in 
    # implemented as Hash Maps. 
    # Create an empty hash map (dictionary) 
    hash_map = {}  
    curr_sum = 0 
    max_len = 0 
    ending_index = -1 
  
    for i in range (0, n): 
        if(arr[i] == 0): 
            arr[i] = -1 
        else
            arr[i] = 1 
  
    # Traverse through the given array 
    for i in range (0, n): 
      
        # Add current element to sum 
        curr_sum = curr_sum + arr[i] 
  
        # To handle sum = 0 at last index 
        if (curr_sum == 0): 
            max_len = i + 1 
            ending_index =
  
        # If this sum is seen before, 
        if curr_sum in hash_map:
              
            # If max_len is smaller than new subarray
            # Update max_len and ending_index
            if max_len < i - hash_map[curr_sum]:
                max_len = i - hash_map[curr_sum]
                ending_index = i
        else
  
            # else put this sum in dictionary 
            hash_map[curr_sum] = i  
          
    for i in range (0, n): 
        if(arr[i] == -1): 
            arr[i] = 0 
        else
            arr[i] = 1 
              
    print (ending_index - max_len + 1, end =" ")
    print ("to", end = " ")
    print (ending_index)
  
    return max_len
  
# Driver Code 
arr = [1, 0, 0, 1, 0, 1, 1
n = len(arr)  
  
maxLen(arr, n) 
      
# This code is contributed 
# by Tarun Garg

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C#

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// C# program to find the largest subarray
// with equal number of 0s and 1s
using System;
using System.Collections.Generic;
  
class LargestSubArray1 {
  
    // Returns largest subarray with
    // equal number of 0s and 1s
    public virtual int maxLen(int[] arr, int n)
    {
  
        // Creates an empty Dictionary hM
        Dictionary<int,
                   int>
            hM = new Dictionary<int,
                                int>();
  
        int sum = 0; // Initialize sum of elements
        int max_len = 0; // Initialize result
        int ending_index = -1;
        int start_index = 0;
  
        for (int i = 0; i < n; i++) {
            arr[i] = (arr[i] == 0) ? -1 : 1;
        }
  
        // Traverse through the given array
        for (int i = 0; i < n; i++) {
            // Add current element to sum
            sum += arr[i];
  
            // To handle sum=0 at last index
            if (sum == 0) {
                max_len = i + 1;
                ending_index = i;
            }
  
            // If this sum is seen before,
            // then update max_len
            // if required
            if (hM.ContainsKey(sum + n)) {
                if (max_len < i - hM[sum + n]) {
                    max_len = i - hM[sum + n];
                    ending_index = i;
                }
            }
  
            else // Else put this sum in hash table
            {
                hM[sum + n] = i;
            }
        }
  
        for (int i = 0; i < n; i++) {
            arr[i] = (arr[i] == -1) ? 0 : 1;
        }
  
        int end = ending_index - max_len + 1;
        Console.WriteLine(end + " to " + ending_index);
  
        return max_len;
    }
  
    // Driver Code
    public static void Main(string[] args)
    {
  
        LargestSubArray1 sub = new LargestSubArray1();
        int[] arr = new int[] {
            1,
            0,
            0,
            1,
            0,
            1,
            1
        };
  
        int n = arr.Length;
        sub.maxLen(arr, n);
    }
}
  
// This code is contributed by Shrikant13

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Output:

 
0 to 5

Complexity Analysis:

  • Time Complexity: O(n).
    As the given array is traversed only once.
  • Auxiliary Space: O(n).
    As hash_map has been used which takes extra space.
  • Thanks to Aashish Barnwal for suggesting this solution.

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